Proof of inequality involving positive number

Question

Let $x$ be a real number such that $x> 0$.

Prove that $x + \frac{1}{4x} \geq 1$.

I tried to do this but cannot prove it.

Answer 1

Use AM-GM inequality

$$x+\frac 1{4x} \ge 2\sqrt{x \cdot \frac 1{4x}}=1$$

Answer 2

Multiplying the equation with $x$ and putting everything on the left side yields $$ 0 \leq x^2 - x +\frac 14 = (x-\frac 12)^2 $$

Answer 3

Completing the square, We've $$x+\frac{1}{4x}=\left(\sqrt x -\frac 1{2\sqrt x}\right)^2+1 \ge 1$$

Answer 4

Two ways.

1) AM-GM. If $x, \frac 1{4x} > 0$ then $\frac{x + \frac 1{4x}}2 \ge \sqrt{x*\frac 1{4x}}$ so $x + \frac 1{4x} \ge 2\sqrt{x*\frac 1{4x}}=1$

2) Completing the square:

Let $x + \frac 1{4x} = k$ then

$x^2 + \frac 14 = kx$

so $x^2 \pm 2*\frac 12 x + \frac 14 = kx \pm x$

$(x \pm \frac 12)^2 = kx \pm x$

$0 \le (x \pm \frac 12)^2 = kx \pm x$

$0 \le (x - \frac 12)^2 = kx - x$

$x \le kx$

$1 \le k = x + \frac 1{4x}$

Or more directly:

$x + \frac 1{4x} = \frac 1x(x^2 + \frac 1{4})$

$= \frac 1x(x^2 -2*\frac 12x + \frac 1{4} + x)$

$=\frac 1x[(x- \frac 12)^2 + x) = \frac 1x(x - \frac 12)^2 + \frac 1x x$

$= \frac 1x(x-\frac 12)^2 + 1 \ge \frac 1x*0 + 1 = 1$>